Explanation

(b)
Since, $f^{''}(x) = 6(x - 1)$
$\Rightarrow$ $f^{'}(x) = 3(x - 1)^{2} + c$ [integrating] …(i)
Also, at the point (2,1) the tangent to graph is $y = 3x - 5$
Slope of tangent=3
$\Rightarrow$ $f^{'}(2) = 3$
$3(2 - 1)^{2} + c = 3$ [from eq. (i)]
$\Rightarrow$ $3 + c = 3$
$\Rightarrow$ $c = 0$
From Eq. (i),
$f^{'}(x) = 3(x - 1)^{2}$
$\Rightarrow$ $f(x) = (x - 1)^{2} + k$ [integrating] …(ii)
Since, it passes through (2,1)
$\therefore$ $1 = (2 - 1)^{2} + k\ \ \ \Rightarrow k = 0$
Hence, equation of function is
$f(x) = (x - 1)^{2}$